Sage,

It might be easier just to back up and look at it from a general view. Back to the EV basics.

EV = (chance of winning)*(amount won) - (chance of losing)*(amount lost)

Since the amount won is the net win after you pay the cost of entry into the game, this can be simplified to:

EV = (chance of winning)*(prize) - (entry cost)

It sounds like in this case the amount of the prize is constant (at least for the sake of argument it is...we know the Expected Value of the prize) and the cost to enter depends on the number of people. Obviously the chance of winning also depends on the number of people.

Using the numbers London Colin chose, which is a $41 prize and a total cost of $12 split evenly between all players, we get the following:

1 player: (1/1)*($41)-($12/1) = $29

2 players: (1/2)*($41)-($12/2) = $14.50 = $29/2

3 players: (1/3)*($41)-($12/3) = $9.67 = $29/3

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12 player: (1/12)*($41)-($12/12) = $2.42 = $29/12

This is of course exactly what London Colin stated.

For a situation where your entry fee is constant regardless of how many people are entered (I'll use a $34 prize and a $5 entry cost), you get this:

1 player: (1/1)*($34)-($5) = $29

2 players (1/2)*($34)-($5) = $12

3 players (1/3)*($34)-($5) = $6.33

4 players (1/4)*($34)-($5) = $3.50

5 players (1/5)*($34)-($5) = $1.80

6 players (1/6)*($34)-($5) = $0.67

7 players (1/7)*($34)-($5) = -$0.14

8 players (1/8)*($34)-($5) = -$0.75

9 players (1/9)*($34)-($5) = -$1.22

10 players (1/10)*($34)-($5) = -$1.60

11 players (1/11)*($34)-($5) = -$1.91

12 players (1/12)*($34)-($5) = -$2.17

This is what Richard Munchkin was describing.

Hopefully this all makes sense to you and I didn't just make it worse.

If not, you can send me an email with some actual numbers and maybe I can explain it better. You don't have to say what game it is if you don't want to, but I think some real numbers would make it easier to explain to you.